Question: The equation of hyperbola $H$ is $\dfrac {(x-2)^{2}}{49}-\dfrac{y^2}{36} = 1$. What are the asymptotes?
Explanation: We want to rewrite the equation in terms of $y$ , so start off by moving the $y$ terms to one side: $\dfrac{y^2}{36} = - 1 + \dfrac {(x-2)^{2}}{49}$ Multiply both sides of the equation by $36$ $y^2 = { - 36 + \dfrac{ (x-2)^{2} \cdot 36 }{49}}$ Take the square root of both sides. $\sqrt{y^2} = \pm \sqrt { - 36 + \dfrac{ (x-2)^{2} \cdot 36 }{49}}$ $ y = \pm \sqrt { - 36 + \dfrac{ (x-2)^{2} \cdot 36 }{49}}$ As $x$ approaches positive or negative infinity, the constant term in the square root matters less and less, so we can just ignore it. $y \approx \pm \sqrt {\dfrac{ (x-2)^{2} \cdot 36 }{49}}$ $y \approx \pm \left(\dfrac{6 \cdot (x - 2)}{7}\right)$ Rewrite as an equality in terms of $y$ to get the equation of the asymptotes: $y = \pm \dfrac{6}{7}(x - 2)+ 0$